Let H he a normal subgroup of a group G and let a G Show tha

Let H he a normal subgroup of a group G. and let a G. Show that (aH)^n = a^nH for all n Z.

Solution

Since H is normal we can consider the quotient group G/H, which has order n.

Fix any a in G. The element aH is in G/H. By Lagrange\'s theorem, the order of the element aH in G/H divides the order of the group G/H, which is n. This means that (aH)ⁿ must be the identity element of G/H, which is H. But from the definition of the operation on G/H we have that (aH)ⁿ = aⁿ H. So we conclude that aⁿ H = H. If we let e denote the identity element of G, we know that e is in H (since H is a subgroup of G) and hence aⁿ = aⁿ * e is in the set aⁿ H (since by definition aⁿ H = {aⁿ * h: h in H}, and e is in H). Since aⁿ is in aⁿ H, and the set aⁿ H is equal to H, we deduce that aⁿ is in H.
Since a was an arbitrary element of G, we deduce that aⁿ is in H for all a in G.